I've turned myself around thinking about indecomposable representations of quivers and Gabriel's Theorem and I would be grateful if someone could point out my misunderstanding.
Suppose you have a $D_4$-like quiver $Q$ with nodes $Q_0 = \{1,2,3,4\}$ and arrows
\begin{equation*}
Q_1 = \left\{ 1\xrightarrow{a}4 \quad 4\xrightarrow{b}2 \quad 4\xrightarrow{c}3 \right\}\,.
\end{equation*}
Let $X$ be the representation of $Q$ where $\dim X_1 = \dim X_2 = \dim X_3 = 1$ and $\dim X_4 = 2$ and, picking a basis at each node, let $X_a = \binom 1 0 $ is the inclusion into the first component, $X_b = (0\,1)$ the projection onto the second component, and $X_c = (1\,1)$. This representation is indecomposable, and by Gabriel's theorem is the unique indecomposable representation with this dimension vector up to isomorphism. But what about the representation $X'$ that is the same as $X$ except we swap the maps $X_b$ and $X_c$? That is $X'_b = (1\,1)$ and $X'_c = (0\,1)$? By symmetry $X'$ will also be indecomposable, but noting that $X_bX_a = 0$ while $X'_aX'_b \neq 0$, these representations are not isomorphic and are not the same up to a change of basis at the nodes.
Where is my misunderstanding here? Is there some "Up to some equivalence" I've overlooked?
Best Answer
The representation $X$ that you describe is not indecomposable.
This becomes clear if you write $X_4$ as the direct sum of the image of $X_a$ and the kernel of $X_c$.